Question

$108\mathrm{g}$ silver (molar mass $108\mathrm{g}{\mathrm{mol}}^{-1}$) is deposited at cathode from ${\mathrm{AgNO}}_3\left(\mathrm{aq}\right)$solution by a certain quantity of electricity. The volume (in $\mathrm{L}$) of oxygen gas produced at $273\mathrm{K}$ and $1\mathrm{bar}$ pressure from water by the same quantity of electricity is:

Answer 1

Step 1: Given data and assumptions:

Weight of silver $\left(\mathrm{Ag}\right)$ deposited at cathode is ${\mathrm{W}}_{\left(\mathrm{Ag}\right)}=108\mathrm{g}$

Molecular weight of Silver is $108\mathrm{g}{\mathrm{mol}}^{-1}$

Temperature is $\mathrm{T}=273\mathrm{K}$

Pressure is $\mathrm{P}=1\mathrm{bar}$

Let the quantity of electricity passed through the system be $\mathrm{Q}$

Volume of oxygen gas produced from water $\left({\mathrm{H}}_2\mathrm{O}\right)$ is ${\mathrm{V}}_{{\mathrm{O}}_2}=\mathrm{V}$

Step 2: Write the Chemical reactions involved during the process:

1. ${\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Ag}$
2. $2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$

The number of electrons involved during the formation of $\mathrm{Ag}$ is ${\mathrm{n}}_{\left(\mathrm{Ag}\right)}=1$

The number of electrons involved during the formation of ${\mathrm{O}}_2$ is ${\mathrm{n}}_{\left({\mathrm{O}}_2\right)}=4$

Step 3: Find the number of moles of ${\mathbf{O}}_{\mathbf{2}}$:

According to Faraday's first law of electrolysis;

1) ${\mathrm{W}}_{\left(\mathrm{Ag}\right)}=\frac{M_{Ag}}{{\mathrm{n}}_{\left(\mathrm{Ag}\right)}\times \mathrm{F}}\times \mathrm{Q}...\left(\mathrm{I}\right)$ where, ${\mathrm{M}}_{\left(\mathrm{Ag}\right)}$ is the molecular weight of ${\mathrm{AgNO}}_3$

Where, ${\mathrm{n}}_{\left(\mathrm{Ag}\right)}=1$ is the number of electrons involved in the formation of $\mathrm{Ag}$

$\mathrm{F}=1\mathrm{Faraday}$

$\mathrm{Q}=$Amount of electricity passed through a system

2) $\mathrm{No}.\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{O}}_2=\frac{1}{{\mathrm{n}}_{\left(O_2\right)}\times \mathrm{F}}\times \mathrm{Q}...\left(\mathrm{II}\right)$

Divide $\mathrm{eq}.\left(\mathrm{I}\right)$ and $\mathrm{eq}.\left(\mathrm{II}\right)$;

$$\begin{gathered} \Rightarrow \frac{{\mathrm{W}}_{\left(\mathrm{Ag}\right)}}{\mathrm{No}.\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{O}}_2}=\frac{{\mathrm{M}}_{\left(\mathrm{Ag}\right)}\times \mathrm{Q}/({\mathrm{n}}_{\left(\mathrm{Ag}\right)}\times \mathrm{F})}{{\displaystyle 1}\times \mathrm{Q}/({\mathrm{n}}_{\left({\mathrm{O}}_2\right)}\times \mathrm{F})} \\ \Rightarrow \frac{108\mathrm{g}}{\mathrm{No}.\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{O}}_2}=\frac{108\mathrm{g}{\mathrm{mol}}^{-1}\times \mathrm{Q}/(\mathrm{n}\times \mathrm{F})}{{\displaystyle 1\times \mathrm{Q}/(4\times \mathrm{F})}} \\ \Rightarrow \mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{O}}_2=\frac{1}{4} \end{gathered}$$

Step 4: Find the volume of ${\mathbf{O}}_{\mathbf{2}}$:

From the Ideal gas equation:

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ \Rightarrow \mathrm{V}=\frac{{\mathrm{n}}_{{\mathrm{O}}_2}\times \mathrm{R}\times \mathrm{T}}{\mathrm{P}} \\ \Rightarrow \mathrm{V}=\frac{{\displaystyle \frac{1}{4}}\mathrm{mol}\times 0.0822\mathrm{L}\mathrm{atm}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}\times 273\mathrm{K}}{1\mathrm{bar}}\left[\because 1\mathrm{atm}=1.01325\mathrm{bar}\right] \\ \Rightarrow \mathrm{V}=\frac{1}{4}\times 0.0822\times 1.01325\times 273 \\ \Rightarrow \mathrm{V}=5.684\mathrm{L} \end{gathered}$$

Hence, the volume of oxygen gas produced at $\mathbf{273}\mathbf{}\mathbf{K}$ and $\mathbf{1}\mathbf{}\mathbf{bar}$ pressure from the water is ${\mathbf{V}}_{{\mathbf{O}}_{\mathbf{2}}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{684}\mathbf{}\mathbf{L}$.